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            <h1 id="seo-header">『算法-ACM竞赛-贪心』HDU 1421 搬寝室 解题报告（超详细）</h1>
            
            
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                <h1 id="『算法-ACM-竞赛-贪心』HDU-1421-搬寝室-解题报告（超详细）"><a href="#『算法-ACM-竞赛-贪心』HDU-1421-搬寝室-解题报告（超详细）" class="headerlink" title="『算法-ACM 竞赛-贪心』HDU 1421 搬寝室 解题报告（超详细）"></a>『算法-ACM 竞赛-贪心』HDU 1421 搬寝室 解题报告（超详细）</h1><p>**搬寝室</p>
<blockquote>
<p>Time Limit: 2000&#x2F;1000 MS Memory Limit: 65536&#x2F;32768 K</p>
</blockquote>
<p><strong>Problem Description</strong><br>搬寝室是很累的,xhd 深有体会.时间追述 2006 年 7 月 9 号,那天 xhd 迫于无奈要从 27 号楼搬到 3 号楼,因为 10 号要封楼了.看着寝室里的 n 件物品,xhd 开始发呆,因为 n 是一个小于 2000 的整数,实在是太多了,于是 xhd 决定随便搬 2<em>k 件过去就行了.但还是会很累,因为 2</em>k 也不小是一个不大于 n 的整数.幸运的是 xhd 根据多年的搬东西的经验发现每搬一次的疲劳度是和左右手的物品的重量差的平方成正比(这里补充一句,xhd 每次搬两件东西,左手一件右手一件).例如 xhd 左手拿重量为 3 的物品,右手拿重量为 6 的物品,则他搬完这次的疲劳度为(6-3)^2 &#x3D; 9.现在可怜的 xhd 希望知道搬完这 2*k 件物品后的最佳状态是怎样的(也就是最低的疲劳度),请告诉他吧.</p>
<p><strong>Input</strong><br>每组输入数据有两行,第一行有两个数 n,k(2&lt;&#x3D;2*k&lt;&#x3D;n&lt;2000).第二行有 n 个整数分别表示 n 件物品的重量(重量是一个小于 2^15 的正整数).</p>
<p><strong>Output</strong><br>对应每组输入数据,输出数据只有一个表示他的最少的疲劳度,每个一行.</p>
<p><strong>Sample Input</strong><br>2 1 1 3</p>
<p><strong>Sample Output</strong><br>4</p>
<p><strong>Author</strong><br>xhd<br><strong>解题思路：</strong><br>题目意思为求 n 个物品，拿 k 对使得消耗的体力最少，或者说是这 k 对物品，每一对中两件物品的质量差平方最小，所以要使得质量差的平方小，只能排序后取质量相邻两个物品作为一对；<br>现在设 dp[i][j]为前 i 件物品组成 k 对所消耗的体力最小；<br>这时分两种情况含有第 i 件物品和不含有第 i 件物品（即第 i 件物品是不是含在第 j 对里） 1.含有 i 件物品 则有 dp[i][j]&#x3D;dp[i-2][j-1]+(ob[i-1]-ob[i])<em>(ob[i-1]-ob[i]) 2.不含第 i 件物品则有 dp[i][j]&#x3D;dp[i-1][j]<br>所以动态转移方程为 ：dp[i][j]&#x3D;min(dp[i-2][j-1]+(ob[i-1]-ob[i])</em>(ob[i-1]-ob[i]),dp[i-1][j]);<br><strong>我们用一组数据来解释状态转移方程。</strong><br><img src="https://img-blog.csdnimg.cn/20190329194151690.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MzYyNzExOA==,size_16,color_FFFFFF,t_70" srcset="/img/loading.gif" lazyload alt="在这里插入图片描述"><br>以最大值 0x3f3f3f3f 为分界线，为什么要使用 0x3f3f3f3f 作为最大值，全是个人习惯，在这里用 0x7f7f7f7f 也可以。但是在有些题目中当两个无穷大相加的时候进位变为负号，成为极小值，为了避免这种情况出现，用 0x3f3f3f3f 作为最大值，比无穷大的一半稍小，但是也是一个极大的数，用来做无穷大最好不过。<br><strong>注 ：</strong></p>
<blockquote>
<p>int 型是 4 个字节 一个字节 8 个位 0x7f7f7f7f 是十六进制 也就是 4 个 0x7f ,一个 0x7f 转化为二进制就是 01111111<br>因为是 int 型 第一个位是符号位 ，因而在 int 型中 0x7f7f7f7f 也就是无穷大的意思。</p>
</blockquote>
<p>以无穷大为界限第一阶梯代表只要当前值可选就选消耗的体力。</p>
<p><img src="https://img-blog.csdnimg.cn/2019032919574534.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MzYyNzExOA==,size_16,color_FFFFFF,t_70" srcset="/img/loading.gif" lazyload alt="在这里插入图片描述"><br>第二阶梯代表有一个数据没有选择的最优消耗体力，他的选择机制是选当前两个，或者跳过去一个，如果跳过这一个比都选要轻松，他会选择跳过当前值，于是这就是状态转移方程，选择机制是选当前值，或者不选当前值。<br>代码如下：</p>
<figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br></pre></td><td class="code"><pre><code class="hljs cpp"><span class="hljs-meta">#<span class="hljs-keyword">include</span><span class="hljs-string">&lt;iostream&gt;</span></span><br><span class="hljs-meta">#<span class="hljs-keyword">include</span><span class="hljs-string">&lt;cstdio&gt;</span></span><br><span class="hljs-meta">#<span class="hljs-keyword">include</span><span class="hljs-string">&lt;cstring&gt;</span></span><br><span class="hljs-meta">#<span class="hljs-keyword">include</span><span class="hljs-string">&lt;cmath&gt;</span></span><br><span class="hljs-meta">#<span class="hljs-keyword">include</span><span class="hljs-string">&lt;algorithm&gt;</span></span><br><span class="hljs-keyword">using</span> <span class="hljs-keyword">namespace</span> std;<br><span class="hljs-type">long</span> <span class="hljs-type">long</span> <span class="hljs-type">int</span> ob[<span class="hljs-number">2001</span>];<br><span class="hljs-type">long</span> <span class="hljs-type">long</span> <span class="hljs-type">int</span> dp[<span class="hljs-number">2001</span>][<span class="hljs-number">2001</span>];<br><span class="hljs-type">const</span> <span class="hljs-type">long</span> <span class="hljs-type">long</span> <span class="hljs-type">int</span> max1=<span class="hljs-number">0x3f3f3f3f</span>;<br><span class="hljs-function"><span class="hljs-type">int</span> <span class="hljs-title">main</span><span class="hljs-params">()</span></span><br><span class="hljs-function"></span>&#123;<br>    <span class="hljs-type">int</span> n,k;<br>    <span class="hljs-keyword">while</span>(cin&gt;&gt;n&gt;&gt;k)<br>    &#123;<br>      <span class="hljs-comment">//  memset(dp,max1,sizeof(dp));  memset 最好用来赋值0或-1，不然会出错。</span><br>        <span class="hljs-built_in">memset</span>(ob,<span class="hljs-number">0</span>,<span class="hljs-built_in">sizeof</span>(ob));<br>        <span class="hljs-keyword">for</span>(<span class="hljs-type">int</span> i=<span class="hljs-number">1</span>;i&lt;=n;i++)<br>            cin&gt;&gt;ob[i];<br>        <span class="hljs-built_in">sort</span>(ob+<span class="hljs-number">1</span>,ob+n+<span class="hljs-number">1</span>);<br>        ob[<span class="hljs-number">0</span>]=<span class="hljs-number">0</span>;<br>        <span class="hljs-keyword">for</span>(<span class="hljs-type">int</span> i=<span class="hljs-number">0</span>;i&lt;=n;i++)<br>        &#123;<br>            <span class="hljs-keyword">for</span>(<span class="hljs-type">int</span> j=<span class="hljs-number">0</span>;j&lt;=n;j++)<br>            &#123;<br>                dp[i][j]=max1;<br><br>            &#125;<br>        &#125;<br>        <span class="hljs-keyword">for</span>(<span class="hljs-type">int</span> i=<span class="hljs-number">0</span>;i&lt;=n;i++) dp[i][<span class="hljs-number">0</span>]=<span class="hljs-number">0</span>;<br>        <span class="hljs-keyword">for</span>(<span class="hljs-type">int</span> i=<span class="hljs-number">2</span>;i&lt;=n+<span class="hljs-number">1</span>;i++)<br>        &#123;<br>            <span class="hljs-keyword">for</span>(<span class="hljs-type">int</span> j=<span class="hljs-number">1</span>;j&lt;=k;j++)<br>                dp[i][j]=<span class="hljs-built_in">min</span>(dp[i<span class="hljs-number">-2</span>][j<span class="hljs-number">-1</span>]+(ob[i<span class="hljs-number">-1</span>]-ob[i])*(ob[i<span class="hljs-number">-1</span>]-ob[i]),dp[i<span class="hljs-number">-1</span>][j]);<br>        &#125;<br>       <span class="hljs-comment">/* for(int i=0;i&lt;n+1;i++)   遇到不会的题目，就按照他的状态转移方程操作后输出dp数组，来研究状态转移方程的实现过程。</span><br><span class="hljs-comment">        &#123;</span><br><span class="hljs-comment"></span><br><span class="hljs-comment">            for(int j=0;j&lt;n+1;j++)</span><br><span class="hljs-comment">            &#123;</span><br><span class="hljs-comment">                cout&lt;&lt;dp[i][j]&lt;&lt;&#x27; &#x27;;</span><br><span class="hljs-comment">            &#125;</span><br><span class="hljs-comment">            cout&lt;&lt;endl;</span><br><span class="hljs-comment">        &#125;*/</span><br>        cout&lt;&lt;dp[n][k]&lt;&lt;endl;<br>    &#125;<br>&#125;<br><br></code></pre></td></tr></table></figure>

                
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